Jun 29 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC Jun 30 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Jul 01 08:00 AM PDT  09:00 AM PDT Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Jul 01 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 17 Aug 2009
Posts: 165

Multiple modulus inequalities
[#permalink]
Show Tags
19 Dec 2009, 02:01
Hi guys
Need to know how to solve multiple modulus inequalities
Ex  Solve x10  2x+3 <= 5x10
I do understand that we need to first find the roots, which happen to be 10, 3/2 and 2 in this case
we can then divide them into intervals 
x<3/2 3/2<x<2 2<x<10 x>10
what do we do post this division into intervals?



Math Expert
Joined: 02 Sep 2009
Posts: 55804

Re: Multiple modulus inequalities
[#permalink]
Show Tags
19 Dec 2009, 03:47
zaarathelab wrote: Hi guys
Need to know how to solve multiple modulus inequalities
Ex  Solve x10  2x+3 <= 5x10
I do understand that we need to first find the roots, which happen to be 10, 3/2 and 2 in this case
we can then divide them into intervals 
x<3/2 3/2<x<2 2<x<10 x>10
what do we do post this division into intervals? If we dot he way you are proposing: Yes, there are 3 check point at which absolute values in the example change their sign: 3/2, 2, and 10 (te values of x for which the expression in  equals to zero). Then you should expand the modulus in these ranges: A. \(x<\frac{3}{2}\) > \(x102x+3\leq{5x10}\) > \((x10)+(2x+3)\leq{(5x10)}\) > \(6x\leq{3}\) > \(x\leq{\frac{1}{2}\) > as we considering the range when \(x<\frac{3}{2}\), then finally we should write \(x<\frac{3}{2}\): (\(\frac{3}{2}\))B. \(\frac{3}{2}\leq{x}\leq{2}\) > \((x10)(2x+3)\leq{(5x10)}\) > \(2x<=3\) > \(x<={\frac{3}{2}\) > \({\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}\); (\(\frac{3}{2}\))(\(\frac{3}{2}\))C. \(2<x<10\) > \((x10)(2x+3)\leq{5x10}\) > \(17\leq{8x}\) > \(x\geq{\frac{17}{8}}\) > \({\frac{17}{8}}\leq{x}<10\); (\(\frac{17}{8}\))(\(10\))D. \(x\geq{10}\) > \((x10)(2x+3)\leq{5x10}\) > \(3\leq{6x\) > \(x\geq{\frac{1}{2}}\) > \(x\geq{10}\); (\(\frac{17}{8}\))(\(10\))After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are: \(x\leq{\frac{3}{2}}\) and \({\frac{17}{8}}\leq{x}\) (\(\frac{3}{2}\))(\(\frac{3}{2}\))(\(\frac{17}{8}\))(\(10\))One important detail when you define the critical points (10, 3/2 and 2) you should include them in either of intervals you check. So when you wrote: x<3/2 3/2<x<2 2<x<10 x>10 You should put equal sign for each of the critical point in either interval. For example: x =<3/2 3/2<x< =2 2<x<1 =0 x>10 Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality.
_________________



Senior Manager
Joined: 12 Mar 2010
Posts: 269
Concentration: Marketing, Entrepreneurship

Re: Multiple modulus inequalities
[#permalink]
Show Tags
01 Feb 2012, 02:31
Could somebody explain this in detail, please?
I didn't get the concept of roots here, it doesn't make the equation zero  isn't it?



Math Expert
Joined: 02 Sep 2009
Posts: 55804

Re: Multiple modulus inequalities
[#permalink]
Show Tags
01 Feb 2012, 02:47
bsaikrishna wrote: Could somebody explain this in detail, please?
I didn't get the concept of roots here, it doesn't make the equation zero  isn't it? When x<=0 then x=x and if x>0 then x=x. For x10: when x<=10 then x10=(x10) (because if x<=10 then x10<=0) and when x>10 then x10=x10 (because if x>10 then x10>0). So, zaarathelab calls "roots" (you can also call "critical point" or "check point") the values of x for which the expression in  equals to zero: in two different cases when x<that value and x>that value the absolute values expands with different sign (it switches the sign at this value). Check Absolute Values chapter of Math Book for more: mathabsolutevaluemodulus86462.htmlHope it helps.
_________________



Manager
Status: May The Force Be With Me (DDAY 15 May 2012)
Joined: 06 Jan 2012
Posts: 198
Location: India
Concentration: General Management, Entrepreneurship

Re: Multiple modulus inequalities
[#permalink]
Show Tags
Updated on: 01 Feb 2012, 03:05
Hi, I solved the modulus by taking the positive & negative scenarios of each modulus. I solved them separately & arrived at 1/2 <= x & x<= 1/2. Am I on the right track or messed up some where??
_________________
Giving +1 kudos is a better way of saying 'Thank You'.
Originally posted by boomtangboy on 01 Feb 2012, 02:55.
Last edited by boomtangboy on 01 Feb 2012, 03:05, edited 1 time in total.



Math Expert
Joined: 02 Sep 2009
Posts: 55804

Re: Multiple modulus inequalities
[#permalink]
Show Tags
01 Feb 2012, 02:59
boomtangboy wrote: Hi,
I solved the modulus by taking the positive & negative scenarios of each modulus. I solved them separately & arrived at 1/2 <= x & x<= 1/2.
Am I on the right track or messed up some where?? The answer to the initial question is x<=3/2 or 17/8<=x, so "1/2 <= x & x<= 1/2" is not right. Check my previous post: you should check different cases for different ranges to get the correct answer.
_________________



Intern
Joined: 25 Sep 2012
Posts: 7
Concentration: General Management

Re: Multiple modulus inequalities
[#permalink]
Show Tags
04 Oct 2012, 11:45
Bunuel wrote: zaarathelab wrote: Hi guys
Need to know how to solve multiple modulus inequalities
Ex  Solve x10  2x+3 <= 5x10
I do understand that we need to first find the roots, which happen to be 10, 3/2 and 2 in this case
we can then divide them into intervals 
x<3/2 3/2<x<2 2<x<10 x>10
what do we do post this division into intervals? If we dot he way you are proposing: Yes, there are 3 check point at which absolute values in the example change their sign: 3/2, 2, and 10 (te values of x for which the expression in  equals to zero). Then you should expand the modulus in these ranges: A. \(x<\frac{3}{2}\) > \(x102x+3\leq{5x10}\) > \((x10)+(2x+3)\leq{(5x10)}\) > \(6x\leq{3}\) > \(x\leq{\frac{1}{2}\) > as we considering the range when \(x<\frac{3}{2}\), then finally we should write \(x<\frac{3}{2}\): (\(\frac{3}{2}\))B. \(\frac{3}{2}\leq{x}\leq{2}\) > \((x10)(2x+3)\leq{(5x10)}\) > \(2x<=3\) > \(x<={\frac{3}{2}\) > \({\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}\); (\(\frac{3}{2}\))(\(\frac{3}{2}\))C. \(2<x<10\) > \((x10)(2x+3)\leq{5x10}\) > \(17\leq{8x}\) > \(x\geq{\frac{17}{8}}\) > \({\frac{17}{8}}\leq{x}<10\); (\(\frac{17}{8}\))(\(10\))D. \(x\geq{10}\) > \((x10)(2x+3)\leq{5x10}\) > \(3\leq{6x\) > \(x\geq{\frac{1}{2}}\) > \(x\geq{10}\); (\(\frac{17}{8}\))(\(10\))After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are: \(x\leq{\frac{3}{2}}\) and \({\frac{17}{8}}\leq{x}\) (\(\frac{3}{2}\))(\(\frac{3}{2}\))(\(\frac{17}{8}\))(\(10\))One important detail when you define the critical points (10, 3/2 and 2) you should include them in either of intervals you check. So when you wrote: x<3/2 3/2<x<2 2<x<10 x>10 You should put equal sign for each of the critical point in either interval. For example: x =<3/2 3/2<x< =2 2<x<1 =0 x>10 Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality. Hi Brunel, How do we decide the signs in front of the modulus in the equation. Eg:: x10  2x+3 <= 5x10......say we are looking in an interval  3/2 to 2. then based on what concentration do we put +or  sign in front of the modulus. and please explain >> step 1 x = sqrt (x^2) ? < step w => x = x ? How do this. please explain concept behind this math. Thanks!!
_________________
 Never Say Die die die die die die die die die 



Math Expert
Joined: 02 Sep 2009
Posts: 55804

Re: Multiple modulus inequalities
[#permalink]
Show Tags
05 Oct 2012, 04:50
Maverick04308 wrote: Hi Brunel, How do we decide the signs in front of the modulus in the equation. Eg:: x10  2x+3 <= 5x10......say we are looking in an interval  3/2 to 2. then based on what concentration do we put +or  sign in front of the modulus. and please explain >> step 1 x = sqrt (x^2) ? < step w => x = x ? How do this. please explain concept behind this math. Thanks!! Explained here: multiplemodulusinequalities88174.html#p1037473Hope it helps.
_________________



Intern
Joined: 26 Dec 2018
Posts: 25
Concentration: Finance, Economics

Multiple modulus inequalities
[#permalink]
Show Tags
11 Jan 2019, 09:49
Bunuel wrote: zaarathelab wrote: Hi guys
Need to know how to solve multiple modulus inequalities
Ex  Solve x10  2x+3 <= 5x10
I do understand that we need to first find the roots, which happen to be 10, 3/2 and 2 in this case
we can then divide them into intervals 
x<3/2 3/2<x<2 2<x<10 x>10
Bun
what do we do post this division into intervals? If we dot he way you are proposing: Yes, there are 3 check point at which absolute values in the example change their sign: 3/2, 2, and 10 (te values of x for which the expression in  equals to zero). Then you should expand the modulus in these ranges: A. \(x<\frac{3}{2}\) > \(x102x+3\leq{5x10}\) > \((x10)+(2x+3)\leq{(5x10)}\) > \(6x\leq{3}\) > \(x\leq{\frac{1}{2}\) > as we considering the range when \(x<\frac{3}{2}\), then finally we should write \(x<\frac{3}{2}\): (\(\frac{3}{2}\))B. \(\frac{3}{2}\leq{x}\leq{2}\) > \((x10)(2x+3)\leq{(5x10)}\) > \(2x<=3\) > \(x<={\frac{3}{2}\) > \({\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}\); (\(\frac{3}{2}\))(\(\frac{3}{2}\))C. \(2<x<10\) > \((x10)(2x+3)\leq{5x10}\) > \(17\leq{8x}\) > \(x\geq{\frac{17}{8}}\) > \({\frac{17}{8}}\leq{x}<10\); (\(\frac{17}{8}\))(\(10\))D. \(x\geq{10}\) > \((x10)(2x+3)\leq{5x10}\) > \(3\leq{6x\) > \(x\geq{\frac{1}{2}}\) > \(x\geq{10}\); (\(\frac{17}{8}\))(\(10\))After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are: \(x\leq{\frac{3}{2}}\) and \({\frac{17}{8}}\leq{x}\) (\(\frac{3}{2}\))(\(\frac{3}{2}\))(\(\frac{17}{8}\))(\(10\))One important detail when you define the critical points (10, 3/2 and 2) you should include them in either of intervals you check. So when you wrote: x<3/2 3/2<x<2 2<x<10 x>10 You should put equal sign for each of the critical point in either interval. For example: x =<3/2 3/2<x< =2 2<x<1 =0 x>10 Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality. Bunuel, I believe that I follow along entirely until you reach the point where you are determining the viability of the solutions. How have you checked the viability of the solutions? In this tutorial:https://gmatclub.com/forum/mathabsolutevaluemodulus86462.html It seems to teach that one must simply determine if the solution is within the range they are checking and if not, then it is not viable. So, in this example, would x<=1/2 satisfy the criteria of x<3/2 because x would have to be less than 1/2 to be less than 3/2? Also, how do you conclude that the inequality does not hold between 3/2 and 17/8? Again, I was able to arrive at the same numbers, but I am uncertain how you figured that the inequality does not hold in that interval. Sorry as I am a bit lost! Thank you in advance for your help! Best, NCH2024



Math Expert
Joined: 02 Sep 2009
Posts: 55804

Re: Multiple modulus inequalities
[#permalink]
Show Tags
12 Jan 2019, 03:55
nch2024 wrote: Bunuel wrote: zaarathelab wrote: Hi guys
Need to know how to solve multiple modulus inequalities
Ex  Solve x10  2x+3 <= 5x10
I do understand that we need to first find the roots, which happen to be 10, 3/2 and 2 in this case
we can then divide them into intervals 
x<3/2 3/2<x<2 2<x<10 x>10
Bun
what do we do post this division into intervals? If we dot he way you are proposing: Yes, there are 3 check point at which absolute values in the example change their sign: 3/2, 2, and 10 (te values of x for which the expression in  equals to zero). Then you should expand the modulus in these ranges: A. \(x<\frac{3}{2}\) > \(x102x+3\leq{5x10}\) > \((x10)+(2x+3)\leq{(5x10)}\) > \(6x\leq{3}\) > \(x\leq{\frac{1}{2}\) > as we considering the range when \(x<\frac{3}{2}\), then finally we should write \(x<\frac{3}{2}\): (\(\frac{3}{2}\))B. \(\frac{3}{2}\leq{x}\leq{2}\) > \((x10)(2x+3)\leq{(5x10)}\) > \(2x<=3\) > \(x<={\frac{3}{2}\) > \({\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}\); (\(\frac{3}{2}\))(\(\frac{3}{2}\))C. \(2<x<10\) > \((x10)(2x+3)\leq{5x10}\) > \(17\leq{8x}\) > \(x\geq{\frac{17}{8}}\) > \({\frac{17}{8}}\leq{x}<10\); (\(\frac{17}{8}\))(\(10\))D. \(x\geq{10}\) > \((x10)(2x+3)\leq{5x10}\) > \(3\leq{6x\) > \(x\geq{\frac{1}{2}}\) > \(x\geq{10}\); (\(\frac{17}{8}\))(\(10\))After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are: \(x\leq{\frac{3}{2}}\) and \({\frac{17}{8}}\leq{x}\) (\(\frac{3}{2}\))(\(\frac{3}{2}\))(\(\frac{17}{8}\))(\(10\))One important detail when you define the critical points (10, 3/2 and 2) you should include them in either of intervals you check. So when you wrote: x<3/2 3/2<x<2 2<x<10 x>10 You should put equal sign for each of the critical point in either interval. For example: x =<3/2 3/2<x< =2 2<x<1 =0 x>10 Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality. Bunuel, I believe that I follow along entirely until you reach the point where you are determining the viability of the solutions. How have you checked the viability of the solutions? In this tutorial:https://gmatclub.com/forum/mathabsolutevaluemodulus86462.html It seems to teach that one must simply determine if the solution is within the range they are checking and if not, then it is not viable. So, in this example, would x<=1/2 satisfy the criteria of x<3/2 because x would have to be less than 1/2 to be less than 3/2? Also, how do you conclude that the inequality does not hold between 3/2 and 17/8? Again, I was able to arrive at the same numbers, but I am uncertain how you figured that the inequality does not hold in that interval. Sorry as I am a bit lost! Thank you in advance for your help! Best, NCH2024 3/2 < 1/2, so x's which are less than 3/2 are also less than 1/2. (\(\frac{3}{2}\))\((\frac{1}{2})\)
_________________



Intern
Joined: 26 Dec 2018
Posts: 25
Concentration: Finance, Economics

Re: Multiple modulus inequalities
[#permalink]
Show Tags
15 Jan 2019, 06:34
Bunuel wrote: Bunuel wrote: Hi guys
Need to know how to solve multiple modulus inequalities
Ex  Solve x10  2x+3 <= 5x10
I do understand that we need to first find the roots, which happen to be 10, 3/2 and 2 in this case
we can then divide them into intervals 
x<3/2 3/2<x<2 2<x<10 x>10
3/2 < 1/2, so x's which are less than 3/2 are also less than 1/2.
(\(\frac{3}{2}\))\((\frac{1}{2})\) [/quote] Thank you, Bunuel!




Re: Multiple modulus inequalities
[#permalink]
15 Jan 2019, 06:34






